Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chmax(T &m, const T &v) {
  if (v > m) return m = v, true;
  return false;
}
template <typename T>
bool chmin(T &m, const T &v) {
  if (v < m) return m = v, true;
  return false;
}
static const int HMAX32 = INT32_MAX / 2;
static const int HMIN32 = -HMAX32;
static const int64_t HMAX64 = INT64_MAX / 2;
static const int64_t HMIN64 = -HMAX64;
void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
static const int maxn = 105;
int arr[maxn][maxn];
int row[maxn], col[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    int rm = INT_MAX;
    for (int j = 1; j <= m; ++j) {
      int t;
      scanf("%d", &t);
      arr[i][j] = t;
      chmin(rm, t);
    }
    row[i] = rm;
    for (int j = 1; j <= m; ++j) {
      arr[i][j] -= rm;
    }
  }
  for (int j = 1; j <= m; ++j) col[j] = arr[1][j];
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (arr[i][j] != arr[1][j]) {
        printf("-1\n");
        return 0;
      }
    }
  }
  if (n > m) {
    int cm = INT_MAX;
    for (int i = 1; i <= n; ++i) chmin(cm, row[i]);
    for (int i = 1; i <= n; ++i) row[i] -= cm;
    for (int j = 1; j <= m; ++j) col[j] += cm;
  }
  int nmov = 0;
  for (int i = 1; i <= n; ++i) nmov += row[i];
  for (int j = 1; j <= m; ++j) nmov += col[j];
  printf("%d\n", nmov);
  for (int i = 1; i <= n; ++i)
    for (int k = 0; k < row[i]; ++k) printf("row %d\n", i);
  for (int j = 1; j <= m; ++j)
    for (int k = 0; k < col[j]; ++k) printf("col %d\n", j);
  return 0;
}